how am I doing it backwards ? 1mV drop becomes 1/81 = 12uV?
What I wrote is the least count of a 22 bit ADC at 5v ref is 5/(2^22 - 1) = 1.2uV which is less than abobe calculated 12uV. So it should read that.(ignoring tolerences and other disturbances).
PS the least count gets multiplied when going from adc to the R1, and voltage across R1 gets divided when coming to ADC.
So 1.2uV makes the least detectable voltage at R1 to be 97.2uV.
Still think I am doing something wrong in my calculation? please correct. thanks.